What is the magnitude of a dual number? I'm not finding information on this.

Question

I'm investigating dual quaternions and am having to learn a lot of stuff myself because I'm finding very few resources on the mathematical background.

I know that the magnitude of a dual quaternion is a dual number. This makes sense because it is a dual number with real quaternions jammed into its real components. Alternately, it is a quaternion with dual numbers jammed into its real components. Both approaches yield the same form.

Question: But what is magnitude of a dual number? Does that question even make sense?

I know that the magnitude of a complex number is the root of sum of the squares of the real numbers. Ditto for a real quaternion. But the dual number operator epsilon (I'll abbreviate it as simply 'e') is not like the irrational unit numbers. Square any of the irrational unit numbers and you get -1. Square e and you get 0. This makes the dual number completely useless as a rotor, but it still has use as a multi-part number. I'm wondering if nature of this dual operator also makes it unsuitable to calculate its magnitude with the Pythagorean Theorem.

Again, the magnitude of a dual quaternion is a dual number, and it doesn't make sense to me to calculate the magnitude again just to get a pure real number, so I've postulated that the magnitude of a dual number is simply itself, like saying that the magnitude of the real number 5 is simply 5. The magnitude of 1189 is 1189. The magnitude of 35 + 14e is 35 + 14e. But I don't know how to prove this or if it is even right.

Help?

Answer 1

The modulus/ norm/ magnitude of a dual number $z=a+b\varepsilon$ is just the absolute value of the real part $|a|$.

Some background: Up to isomorphism, there are exactly $3$ unital $2$-dimensional algebras over $\Bbb R$. They are the complex numbers, split-complex numbers (AKA hyperbolic numbers), and dual numbers. All of them are defined to have the same formulae for conjugation and modulus. For any member of one of these algebras $z=a+b\omega$, where $\omega$ is the "imaginary" unit of the given algebra, the conjugate is defined as $\overline z = a-b\omega$ and the modulus is defined by $\|z\|^2 = z\overline z = \overline z z$.

We can see that this is a good definition, because the norm will always give us a member of $\Bbb R$: $\|z\| = \sqrt{z\overline z} = \sqrt{(a+b\omega)(a-b\omega)} = \sqrt{a^2-b^2\omega^2}$, where $\omega^2 \in \Bbb R$.

In the case of the dual numbers, $\varepsilon$ is defined by $\varepsilon \ne 0$ and $\varepsilon^2 = 0$. Then for $z=a+b\varepsilon$, we have $\|z\| = \sqrt{a^2-b^2\varepsilon^2} = \sqrt{a^2} = |a|$.

NOTE: You might also be interested to know that each of these algebras has a well-defined exponential form as well: $z=a+b\omega = \pm \|z\|e^{\operatorname{Arg}(z)\omega}$, where $\operatorname{Arg}(z)$ is akin to an angle -- in fact in the complex numbers it IS the angle. The norm of the dual number $re^{\theta\varepsilon}$ is just $|r|$.

In fact, the exponential form of the dual numbers is particularly easy. We define the exponential of a number by the infinite series $$e^x = \sum_{i=0}^\infty \frac {x^i}{i!}$$

In the case of dual numbers we have $e^{\theta\varepsilon} = 1 + \theta\varepsilon + \frac {(\theta\varepsilon)^2}2 + \frac {(\theta\varepsilon)^3}6+\cdots$. Notice that all of the terms which include a power of $\varepsilon$ greater than $1$ are $0$. Therefore this infinite series has only $2$ non-zero terms: $e^{\theta\varepsilon} = 1 + \theta\varepsilon$. Therefore the exponential form of the dual number $z=a+b\varepsilon$ is just $$z=ae^{(b/ a)\varepsilon}$$

The cavaet in this representation is clear, however: $\operatorname{Arg}(z)=\theta$ does not exist for pure imaginary numbers in the dual number plane.